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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 3 of 120

Marks: +1, -0

If

\log_{10} 2, \log_{10}(2^{x}-1), \log_{10}(2^{x}+3)

are in AP, then what is

x

0
1
$\log_{2} 5$
$\log_{5} 2$

Solution: 👈: Video Solution

(c) Given that,

\log_{10} 2, \log_{10}(2^{x}-1)

,

\log_{10}(2^{x}+3)

are in AP.

\therefore 2 \log_{10}(2^{x}-1)=\log_{10} 2+\log_{10}(2^{x}+3)

\log_{10}(2^{x}-1)^{2}=\log_{10} 2(2^{x}+3)

\Rightarrow 2^{2x}+1-2 \cdot 2^{x}=2 \cdot 2^{x}+6

\Rightarrow (2^{x})^{2}-4(2^{x})-5=0

Let

\; \; 2^{x}=y

\Rightarrow \; \; y^{2}-4y-5=0

\Rightarrow \; \; y=5

or

y=-1

(Ignore because

2^{x}

cannot be negative)

\Rightarrow y=5 \Rightarrow 2^{x}=5

x=\log_{2} 5

Hence, option (c) is correct.

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